Question

If $f\left(x\right)={\sin }^{2}x+{\sin }^2(x+\frac{\pi }{3})+\cos x\cos (x+\frac{\pi }{3})$ and $g\left(\frac{5}{4}\right)=1,$ then $gof\left(x\right)$ is equal to

Answer 1

Given $f\left(x\right)={\sin }^{2}x+{\sin }^2(x+\frac{\pi }{3})+\cos x\cos (x+\frac{\pi }{3})$

and $g\left(\frac{5}{4}\right)=1$

$f\left(x\right)=\frac{1}{2}[2{\sin }^{2}x+2{\sin }^2(x+\frac{\pi }{3})+2\cos x\cos (x+\frac{\pi }{3})]$

$=\frac{1}{2}[1-\cos 2x+1-\cos (2x+\frac{2\pi }{3})+\cos (2x+\frac{\pi }{3})+\cos \frac{\pi }{3}]$

$=\frac{1}{2}[\frac{5}{2}-\cos 2x-\cos (2x+\frac{2\pi }{3})+\cos (2x+\frac{\pi }{3})]$

$=\frac{1}{2}[\frac{5}{2}-2\cos (2x+\frac{\pi }{3})\cos \left(\frac{\pi }{3}\right)+\cos (2x+\frac{\pi }{3})]$

$=\frac{1}{2}[\frac{5}{2}-\cos (2x+\frac{\pi }{3})+\cos (2x+\frac{\pi }{3})]=\frac{5}{4}$

$f\left(x\right)=\frac{5}{4}$ for all $x\in R$

Therefore for any $x\in R$ we have

$gof\left(x\right)=g\left(f\left(x\right)\right)=g\left(\frac{5}{4}\right)=1$

Thus $gof\left(x\right)=1forallx\in R$

Hence $gof\left(x\right)=1$